Question

我正在尝试让我的程序从字符串中相邻的列表中检查2个字符的实例的字符串，并返回另一个字符串以替换这两个字符。

def main（）：

dubs = ["ai", "ae", "ao", "au", "ei", "eu", "iu", "oi", "ou", "ui"]
newdubs = [ "eye", "eye", "ow", "ow", "ay","eh-oo", "ew", "oy", "ow","ooey"]

word = input("ENTER WORD : " )
count = 0
fin = []
while count < len(word):

   if word[count:count+2] in dubs:

        if word[count:count+2] == dubs[0]:
            fin.append(newDubs[0] + "-")

        if word[count:count+2] == dubs[1]:
            fin.append(newDubs[1] + "-")

        if word[count:count+2] == dubs[2]:

            fin.append(newDubs[2] + "-")

        if word[count:count+2] == dubs[3]:
            fin.append(newDubs[3] + "-")

        if word[count:count+2] == dubs[4]:
            fin.append(newDubs[4] + "-")

        if word[count:count+2] == dubs[5]:
            fin.append(newDubs[5] + "-")

        if word[count:count+2] == dubs[6]:
            fin.append(newDubs[6] + "-")

        if word[count:count+2] == dubs[7]:
            fin.append(newDubs[7] + "-")

        if word[count:count+2] == dubs[8]:
            fin.append(newDubs[8] + "-")

       if word[count:count+2] == dubs[9]:
            fin.append(newDubs[9] + "-")

    if word[count:count+2] not in dubs:
        fin.append(word[count])

    count+=1
fin= "".join(fin)

print(fin)

像wanai这样的词，我希望wan-eye 结果是waneye-i
我还需要进行检查，以查看dubs之前的字符是否是元音，但不必担心，直到它能正常工作为止

Answer 1

使用class MobileBind extends React.Component { constructor(props) { super(props); this.state = { birthYear: '', defaultYear: true, }; } render() { const { birthYear, } = this.state; const createNumberSet = (min = 0, max = 10) => { if (min > max) return []; return Array(max - min + 1) .fill(min) .map((value, index) => value + index) .filter(value => value >= min); }; const currentYear = moment().year(); const years = createNumberSet(currentYear - 100, currentYear); const months = createNumberSet(1, 12); const days = createNumberSet(1, 31); return ( <div className={style['birth-date-input']} style={{ margin: 0 }}> <FormControl> <Select name="birthYear" defaultValue={birthYear} onChange={this.handleChange} onClick={this.handleDefalutYear} native> { [ <option key="birthYear-default" value="" disabled > Year</option>, ...(years || []).map(value => <option key={`birthYear-${value}`} value={value} >{value}</option>) ] } </Select> </FormControl> </div> } handleChange = event => { const target = event.target; const value = target.type === 'checkbox' ? target.checked : target.value; const name = target.name; this.setState({ [name]: value }); } handleDefalutYear = () => { this.setState({defaultYear:false}); } + zip()：

replace()

Answer 2

我将重组您的代码以使其更具模块化：

dubs = ["ai", "ae", "ao", "au", "ei", "eu", "iu", "oi", "ou", "ui"]
newdubs = [ "eye", "eye", "ow", "ow", "ay","eh-oo", "ew", "oy", "ow","ooey"]

def dubbizer(word):
   for itter in range(len(dubs)):
       word = word.replace(dubs[itter], "-"+newdubs[itter])
   return word

print(dubbizer("wanai"))

这应该为您提供wan-eye

的输出

不替换：

dubs = ["ai", "ae", "ao", "au", "ei", "eu", "iu", "oi", "ou", "ui"]
newdubs = [ "eye", "eye", "ow", "ow", "ay","eh-oo", "ew", "oy", "ow","ooey"]

def dubbizer(word):
   for itter in range(len(dubs)):
     while dubs[itter] in word:
       word = word[:word.find(dubs[itter])]+"-"+newdubs[itter]+word[word.find(dubs[itter])+len(dubs[itter]):]
   return word

print(dubbizer("wanai"))

修复程序，该程序检查单词彼此相邻的2个字符的出现

2 个答案: